NATIONAL SENI CERTIFICATE GRADE MATHEMATICS P3 NOVEMBER 0 MEMANDUM MARKS: 00 This memorandum consists of 6 pages.
Mathematics/P3 DBE/November 0 NOTE: If a candidate answered a question TWICE, mark the FIRST attempt ONLY. If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed out question. Consistent accuracy applies in ALL aspects of the memorandum. QUESTION. The number of times the CD was played. Afrikaans: Getalkerewat die CD gespeel is. (). 5,000 4,500 4,000 3,500 3,000,500 Scatter plot showing the number of times a CD was played vs the CD sales in the following week all 0 points plotted correctly marks if 5 9 points are plotted correctly mark if 4 points are plotted correctly. (3),000,500,000 500 0 0 5 0 5 0 5 30 35 40 45 50 55.3 a 93,06 (93,057554...) b 74,8 (74,8057554...) yˆ 93,06 + 74, 8x calculating a and b equation.4 r 0,95 (0,945885...).5 yˆ 93,06 + 74,8(45) 3635,66 3635 3650 (to the nearest 50) Note: Penalise mark for incorrect rounding off. substitution () ().6 There is a very strong positive relationship between the number of times that a CD was played and the sales of that CD in the following week. strong () [3]
Mathematics/P3 3 DBE/November 0 QUESTION. Yes. The events Pass and Fail are mutually exclusive. It is not possible for pass and fail to take place at the same time. There is no intersection between the two sets. P(Pass and Fail) 0 P(Pass) 0,59 P(Fail) 0,4 P(Pass) + P(Fail) 0,59 + 0,4 Note: If a candidate answers No then award 0 marks P(Pass and Fail) 0 / No intersection of the sets The events Pass and Fail are mutually exclusive. Yes P(Pass and Fail) 0 / no intersection between the sets. () Yes P(Pass and Fail) 0 / No intersection between the sets (). Afrikaans Ja. Die gebeurtenisse Slaag en Druip is onderling uitsluitend. Dit is nie moontlik dat slaag en druip gelyktydig plaasvind nie. P(Slaag en Druip) 0 PASS FAIL TOTAL Males 46 3 78 Females 7 50 Total 8 8 00 78 P ( Male) 0,39 Note: 00 If a candidate answers No 8 P ( Pass) 0,59 then award 0 marks 00 46 P ( Male and Pass) 0,3 00 P(Male) P(Pass) 0,39 0,59 0,3 (0,30) Ja P(Slaag en Druip) 0 / geen snyding () 78 P ( Male) 0, 39 or 00 8 P ( Pass) 0,59 00 P ( Male and Pass) 0, 3 P ( Male) P(Pass) 0, 3 P ( Male) P(Pass) P(Male and Pass) Passing the competency test is independent of gender. conclusion
Mathematics/P3 4 DBE/November 0 P ( Female) 0,6 00 8 P ( Pass) 0,59 00 7 P ( Female and Pass) 0,36 00 P(Female) P(Pass) 0,6 0,59 0,36 (0,3599) P ( Female) P(Pass) P(Female and Pass) Passing the competency test is independent of gender. 0 P ( Female) 0, 6 or 00 8 P ( Pass) 0,59 00 P ( Female and Pass) 0, 36 P ( Female) P(Pass) 0, 36 conclusion [6]
Mathematics/P3 5 DBE/November 0 QUESTION 3 3. 0 5 Histogram showing the frequency of the lifespan of a television (years) 8 7 intervals 3 bars correct 6 bars correct (3) 0 5 6 5 Note If the candidate draws a bar graph, award max marks 3. 0 0 4,95 5,65 6,35 7,05 7,75 8,45 9,5 Lifespan (in years) Frequency Midpoint 4,95 x< 5,65 5,3 5,65 x< 6,35 6 6 6,35 x< 7,05 8 6,7 7,05 x< 7,75 7 7,4 7,75 x< 8,45 5 8, 8,45 x< 9,5 8,8 0 Note: If candidate works out average ( x ) of midpoints, answer is 7,05 then 0 marks 5,3 + 6 6 + 8 6,7 + 7 7,4 + 5 8, + 8,8 x 50 35, 50 7,0 ( x 7,0) years frequencies midpoints 50 (3) 3.3 The required area is 98% to the right of some value. This value is at standard deviations on the left of the mean. x σ 7,0 (0,76) 5,5 years x σ 7,0 (0,76) (3)
Mathematics/P3 6 DBE/November 0 3.4 They can issue a 5-year guarantee. The average lifespan of a set is 7,0 years - which is in excess of 5 years. 98% of the sets lasted for more than 5,5 years. Very few sets have lasted less than 5 years. The number of sets of this brand that will be returnedshould be minimal if a 5-year guarantee is issued. Afrikaans Hullekan n 5 jaar-waarborguitreik. Die gemiddelde lewens duur van 'n televisiestel is 7,0 jaar wat 5 jaar oorskry. 98% van die stelle het langer as 5,5jaargehou. 'n Klein aantal stelle het vir minder as 5 jaar gehou. Die aantal stele wat terug geneem sal moet word sal minimal wees indien 'n 5 jaarwaarborg uitgereik word. Issue the 5-year guarantee reason () kan n 5 jaarwaarborg uitreik rede () []
Mathematics/P3 7 DBE/November 0 QUESTION 4 4. Sunny 4 / 7 Rainy 3 / 7 Cycle Drive Train 7/ 0 / 5 / 0 Cycle Drive / 9 5 / 9 Outcome(Sunny, cycle) Outcome(Sunny, drive) Outcome(Sunny, train) Outcome(Rainy, cycle) Outcome(Rainy, drive) Sunny branch Rainybranch cycle, drive, train branches on both weather types probabilities listed outcomes listed (5) Train / 3 Outcome(Rainy, train) Cycle 0,7 Outcome(Sunny, cycle) Sunny 0,5748... Drive Train 0, 0, Outcome(Sunny, drive) Outcome(Sunny, train) Rainy 0,4857... Cycle Drive 0,... 0,5555... Outcome(Rainy, cycle) Outcome(Rainy, drive) Train 0,33333... Outcome(Rainy, train)
Mathematics/P3 8 DBE/November 0 4.. P(Rainy, Cycle) 3 7 9 P(Rainy, Cycle) 0,48... 0,... 0,047690476 0,05 Note: 3 If + then 0 marks 7 9 3 7 9 in any form (must be from multiplication) () or 4,76% 4.. P(Train) P(Train) 4 3 0, + 7 7 3 5 0, 0% 4 3 0, + 4 3 0, and 7 7 3 7 7 3 0,0574... + 0,48... addition (in any form) 5 (3) 0, 0% 4.3 4 3 5 P(Drive) 0, + 7 7 9 37 05 0,3538... 37 Vusi drives for 45 87 days (86,333...) 05 Accept: 86 days 4 3 5 P(Drive) 0, 45 + 45 7 7 9 8 + 58,333 87 days (86,333...) Accept: 86 days 4 3 5 0, and 7 7 9 addition 37 05 4 3 5 0, and 7 7 9 addition 8 + 58, 333 [4]
Mathematics/P3 9 DBE/November 0 QUESTION 5 5.. Number of PIN codes 0 0 0 0 0 0 5 00 000 5.. Number of PIN codes 0 9 8 7 6 30 40 Number of PIN codes 0! 5! 30 40 5. Number of PINs that DO NOT contain 9s 9 9 9 9 9 59 049 P(at least one 9) P(no 9s) 59049 00000 0,4 Number of PINs that DO NOT contain 9s 9 9 9 9 9 59 049 Number of PINs that contain AT LEAST one 9 00 000 59 049 40 95 P(at least one 9) 4095 00000 0,4 0 multiplication 0! 5! 9 59 049 9 59 049 4095 59049 00000 () () () [8]
Mathematics/P3 0 DBE/November 0 QUESTION 6 6. T T 3 where T, k k+ k + T k where T, k + k+ T k + T T k+ k + T k T 3 k + k+ T k + T k 6. T T T T where T, T 5, k ( k+ k ) k+ + k+ 4 7 3 4 44 + + 3 + 3 4 + 4 The next term of the sequence is 5 44 + + 5 8 T ( T T ) T k+ k+ k + k+ T T 5 k answer 4 7 3 4 44? 4 7 3 4 44 79 3 6 0 35 3 5 9 5 4 6 The next term of the sequence is 79. answer Note: This sequence can be represented by the following recursive formula: 3 Tn+ Tn + n n n 3 + 3 where T 4 and n () [6]
Mathematics/P3 DBE/November 0 QUESTION 7 L F P Q M N 7. Draw a point P on FG such that FP LM and a point Q on FH such that FQ LN. Note: No construction constitutes a breakdown, hence no marks In FPQ and LMN. Fˆ Lˆ (given). FP LM (construction) 3. FQ LN (construction) FPQ LMN (SAS) F PˆQ LMˆ N ( s) But F ĜH LMˆ N (given) F PˆQ FĜH G construction All three statements must be given FPQ LMN (SAS) F PˆQ LMˆ N F PˆQ FĜH H PQ GH (corresponding angles ) FP FQ FG FH (PQ GH ; Prop Th) LM LN FG FH PQ GH FP FQ FG FH (7)
Mathematics/P3 DBE/November 0 V x 0 x 0 4 T P 6 9 9 K R 7. VP PR x VT TK 0 4 9 6 x 0 6 x 6 x 8 (PT RK;Prop Th) VP PR VT TK (PT RK; Prop Th) substitution VP VT (PT RK; Prop Th) VR VK x 0 4 x 0 0x 00 8x 4 x 96 x 8 VP VR VT VK (PT RK; Prop Th) substitution []
Mathematics/P3 3 DBE/November 0 QUESTION 8 8.... equal to the angle subtended by the chord in the alternate segment. 8. R () M Q 4 76 9 9 7 7 63 3 75 c 34 b 76 P a 9 34 4 05 d T U a 9 (tan ch.thm) Q PˆR 34 ( s in same seg) c 4 b 76 (adj s on str. line) Qˆ 76 ( s in same seg) d 05 (ext cyclic quad) a 9 (tan ch. thm) Tˆ c (tan ch. thm) c + 34 75 (tan ch. thm) c 4 b 76 (adj s on str. line) W a 9 tan ch. thm QPˆR 34 s in same seg c 4 b 76 Qˆ 76 d 05 ext cyclic quad a 9 tan ch. thm Tˆ c tan ch. thm c + 34 75 tan ch. thm c 4 b 76 d 05 (9) (9) d 05 (adj s on str. line) An alternative solution for calculating d: Qˆ RPˆT 76 ( s in same seg) d + R Tˆ Q PQˆ T + QPˆR + RPˆT (ext ) d + 34 9 + 34 + 76 d 05 [0]
Mathematics/P3 4 DBE/November 0 QUESTION 9 C x 360 x O x B 4 x 3 x 80 x K T 3 A 9. A ÔB x ( circ centre circumference) Tˆ 80 x (opp cyclic quad suppl) 9. C ÂT x ( sum ) Kˆ x (ext cyclic quad) C ÂT Kˆ BK AC (corresponding s ) Kˆ Ĉ x (ext cyclic quad) Bˆ 4 x ( sum ) Bˆ 4 Ĉ x BK CA (corresponding s ) C ÂT x ( sum ) B Kˆ A 80 x (opp cyclic quad) C ÂT + BKˆ A 80 BK AC (coint s supp) A ÔB x circ centre circumference opp cyclic quad suppl (3) C ÂT x sum Kˆ x ext cyclic quad corresponding s (5) Kˆ Ĉ x ext cyclic quad Bˆ 4 x sum corresponding s (5) C ÂT x sum B Kˆ A 80 x opp cyclic quad co-int s supp (5)
Mathematics/P3 5 DBE/November 0 9.3 In BKT and CAT. C ÂT Kˆ ( x). Tˆ is common 3. A ĈT Bˆ 4 ( sum ) BKT CAT ( ) 9.4 AC AT ( s) KB KT AC 7 KB C ÂT Kˆ Tˆ is common AC KB s AT KT (3) (3) [4]
Mathematics/P3 6 DBE/November 0 QUESTION 0 B D 6,5x O,5x M 4x C A 0. DC 3x CD 3 x 0. 3 OD x 5 OM x 0.3 BO OD (radii) AM MB units (line from circ cent ch) 5 3 + x x (Pythagoras) 5x 69x 44 + 4 4 44x 44 4 x 4 x ± x ( x > 0) OD 3 x MB 5 3 + x x or + 6,5x 4,5x 5 69 or + x x 4 4 () () 3 The radius ( ) 3 units. [7]